Question

Butyric acid is responsible for the foul smell of rancid butter. The \(\mathrm{pK}_{a}\) of butyric acid is \(4.84\). (a) Calculate the \(\mathrm{pK}_{b}\) for the butyrate ion. (b) Calculate the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of butyric acid. (c) Calculate the pH of a \(0.050 \mathrm{M}\) solution of sodium butyrate.

Short answer

(a) The pKb for the butyrate ion is 9.16. (b) The pH of a 0.050 M solution of butyric acid is 2.46. (c) The pH of a 0.050 M solution of sodium butyrate is 8.34.

Step-by-step solution

(a) Calculate the pKb for the butyrate ion.

Given the pKa of butyric acid as 4.84, we can calculate the pKb for the butyrate ion using the relationship pKa + pKb = 14. Therefore, pKb = 14 - pKa = 14 - 4.84 = 9.16.

(b) Calculate the pH of a 0.050 M solution of butyric acid.

To calculate the pH of a 0.050 M solution of butyric acid, we start by setting up the Ka expression for the weak acid dissociation: \[K_{a} = \frac{[\text{H}^{+}][\text{Butyrate}^{-}]}{[\text{Butyric Acid}]}\] For simplicity, call the dissociated amount of butyric acid as x. Then, we can write: \[K_{a} = \frac{(x)(x)}{(0.050 - x)}\] Since the pKa of butyric acid is 4.84, we can find Ka as follows: \[K_{a} = 10^{-\text{pKa}} = 10^{-4.84}\] Plug in the Ka value and solve for x (assuming x << 0.050): \[(10^{-4.84}) = \frac{(x)(x)}{(0.050)}\] Solving for x, we get the [H+] concentration: \[[\text{H}^{+}] = x = 3.48 \times 10^{-3}\] Now, we can calculate the pH using the formula: \[\text{pH} = -\log[\text{H}^{+}] = -\log(3.48 \times 10^{-3}) = 2.46\] So, the pH of a 0.050 M solution of butyric acid is 2.46.

(c) Calculate the pH of a 0.050 M solution of sodium butyrate.

For this part, we'll use the pKb and set up the Kb expression for the butyrate ion dissociation as follows: \[K_{b} = \frac{[\text{OH}^{-}][\text{Butyric Acid}]}{[\text{Butyrate}^{-}]}\] As we did for part (b), call the dissociated amount of butyrate ion as y. Then, we can write: \[K_{b} = \frac{(y)(y)}{(0.050 - y)}\] Since the pKb of the butyrate ion is 9.16, we can calculate Kb as follows: \[K_{b} = 10^{-\text{pKb}} = 10^{-9.16}\] Plug in the Kb value and solve for y (assuming y << 0.050): \[(10^{-9.16}) = \frac{(y)(y)}{(0.050)}\] Solving for y, we get the [OH-] concentration: \[[\text{OH}^{-}] = y = 2.18 \times 10^{-6}\] Now, we can calculate the pOH using the formula: \[\text{pOH} = -\log[\text{OH}^{-}] = -\log(2.18 \times 10^{-6}) = 5.66\] Recall that pH + pOH = 14. Therefore, we can calculate the pH of the sodium butyrate solution as follows: \[\text{pH} = 14 - \text{pOH} = 14 - 5.66 = 8.34\] So, the pH of a 0.050 M solution of sodium butyrate is 8.34.